The Mean Value Theorem
The Fundamental Theorem, Part 1
The Fundamental Theorem, Part 2

The Mean Value Theorem
A continuous function on a closed interval assumes its average value
at least once in the interval. More precisely, if

is continuos on
, then there is at least one point  in 
such that
.

The theorem is not true if the function is not continuous in the closed interval.
A discontinuous function need not assume its average value.

For the function
    for            Figure 4.15, page 353.
                       
There is no point   in the domain of the function at which
,  the average value of the function.

See Example 1 - 2, pages 352 - 353.


Definite integrals can be used to define functions.

defines
 as a function of the upper limit of integration  .
The value of the function depends on the value of the upper limit of integration.

Example
1]


Note that
.

is defined in terms of a definite integral with the independent variable
appearing as the upper limit of integration. When we take the derivative of
with respect to , we get the integrand with the variable of integration
  replaced by  .

In fact, this result is generally true!
The Fundamental Theorem of Calculus, Part 1
If

is a continuous function on
, then the function

has a derivative at every point   in   and

.

We can put this result in another way:
For every continuous function
,
the differential equation
has a solution.
The solution can be written as a definite integral.

Example
2]
.
Though we do not know how to evaluate the definite integral, we can write down
its derivative.


3] Using the Chain Rule.
If
,
let , so we can use the Chain Rule:







See Examples 3 - 6, pages 354 - 356.

Now we will evaluate definite integrals using antiderivatives.
The Fundamental Theorem of Calculus, Part 2
If    is a continuous function on     and if
    is any antiderivative of   on  ,  then
.

This result often makes the calculation of definite integrals much easier.
If we know the antiderivative, we do not have to calculate the limit of
Riemann sums.

Example
4]
Find
.
Using the definition of a definite integral as a Riemann Sum,
we would have to calculate
,
which would be not too easy.
But we know the antiderivative

so


                   

                   .
Much easier!!


See Examples 7 - 9, pages 359 - 361.


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